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A border of uniform width is placed around a rectangular [#permalink]

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20 Dec 2012, 05:33

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A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches. If the area of the border is 144 square inches, what is the width of the border, in inches?

A border of uniform width is placed around a rectangular photograph that measures 8 inches by 10 inches. If the area of the border is 144 square inches, what is the width of the border, in inches?

(A) 3 (B) 4 (C) 6 (D) 8 (E) 9

Consider the diagram below:

Attachment:

photograph.png [ 5.15 KiB | Viewed 51616 times ]

The area of just the photograph is 8*10=80 square inches.

The area of the photograph with the border is \((10+2x)(8+2x)=4x^2+36x+80\) square inches.

The difference is 144 square inches, thus \((4x^2+36x+80)-80=144\) --> \(4x^2+36x-144=0\) --> \(x^2+9x-36=0\) --> \((x-3)(x+12)=0\) --> \(x=3\).

Re: A border of uniform width is placed around a rectangular [#permalink]

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02 Jan 2015, 18:03

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Length of Picture L1 = 10 Width of Picture W1 = 8

Area of Picture = L1 X W1 = 80

Total Area including frame = L2 X W2 = 144

So L2 and W2 should be multiples/factors of 144.

Prime factorization of 144 = 2 X 7 X 11

Only multiples that make any sense are 14 and 11. (2 x 77 is impossible and so is 22 x 7; total length/width can't be less than length/width of the photograph).

So length (L2) should be 14 and Width (W2) should be 11.

Re: A border of uniform width is placed around a rectangular [#permalink]

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19 Nov 2015, 00:52

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I found substitution as an easier solution to this problem.

Given that photograph dimensions are 8 X 10. Its area is 80. Total area(frame + photo) - area of photo = 144 (8+2x) (10+2x) - 80 = 144 --> Here x is the border width. Adding the border on both sides, extra padding is 2x. (8+2x) (10+2x) = 224

Solving this will give a quadratic equation, which will consume some time to get to. Easier would be to substitute values.

Substitute smallest option, i.e A. Fits right in!! Answer Choice A is correct!
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Re: A border of uniform width is placed around a rectangular [#permalink]

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25 Aug 2014, 22:19

Once we get the idea that the width (w) is to be added twice to 10 and 8 we can find by substitution This how I did the problem Area of frame = 80 Area of photograph = 144 Total Area = 224

Now we know the new area will be (8+2w)*(10+2w) Substitute option A for w (8+6)*(10+6) = 14*16 = 224

Re: A border of uniform width is placed around a rectangular [#permalink]

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24 Apr 2015, 12:40

I didn't find this question difficult, however I still managed to make a mistake. I basically thought that the outer size of the board matches with the edges of the picture. This would give the following equation: 144= 2(8x) +2x(10-2x). However , the inner sides of the board match with the picture. I think that question did not state it clear enough.
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A border of uniform width is placed around a rectangular [#permalink]

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18 May 2015, 03:22

There is actually quite a nifty little way of doing this. For those who struggle with the whole quadratic solution.

You can do the following:

Since a uniform width is placed around it. We need to find a number where A remains 2 bigger than B. (Since just as much is added to the length as to the width. A will always be 2 more than B)

Since area is 224. We need to find a number for A that is 2 greater than B. You might ask yourself well how am I supposed to do that?

If you can realize that 224 is 1 less than a perfect square. You can use the following: \((x+y)(x-y)\), which in this case is \((15+1)(15-1)\). So we got a = 16 and b = 14.

Now, since the inner border is 8. and we add 2x we got 8 + 2x = 14. x = 3.

This might seem like a weird approach, but if you can quickly realize the relation here, you can solve it in less than a minute

A border of uniform width is placed around a rectangular [#permalink]

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29 Nov 2015, 07:39

Given Info: We are given a rectangular photograph of size 8X10 inches. The photograph is surrounded by a uniform width border whose area is 144 square inches.

Interpreting the Problem: In order to find the width of the border, we need to first assume the width of the border in terms of x, and then use the relation of the difference of the areas of 'Photograph+Border' and 'Photograph' to calculate the width of the border.

Solution: Let the width of the border be a.

Then the width of the larger rectangle formed as indicated in the figure will be 8+2a (Uniform width on both sides) The length of the larger rectangle formed as indicated in the figure will be 10+2a (Unifrom width on both sides)

This is shown in the figure

Attachment:

6.png [ 8.39 KiB | Viewed 29765 times ]

Now calculating areas

Area of the larger rectangle = (8+2a)*(10+2a) = \(4a^2+36a+80\) Area of the photograph = 8*10 = 80

Now the difference of the areas of larger rectangle and photograph is given to us as 144 square inches.

Forming an equation based on above information

Area of larger rectangle- Area of the photograph = \(4a^2+36a+80\)-\(80\)=144 \(4a^2+36a\)=\(144\) \(4a^2+36a-144=0\) \(a^2+9a-36=0\) \(a^2+12a-3a-36=0\) \((a-3)(a+12)=0\)

Only solution valid for this quadratic equation is a=3 (Because the length of the border of photograph cannot be negative)

So the width of the border will be equal to 3 inches.

I'd like to ask you the main question I have: if the photograph has sides 8 x 10, the border should not have a larger-than-photograph size?

You mean the total or final dimensions of picture with border? If that is the case then yes, you are correct and that is what is mentioned in a-border-of-uniform-width-is-placed-around-a-rectangular-144434.html#p1158913 with dimensions of the photograph = 8 x 10 while those of the entire thing (photograph+border) = 14 x 16.
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Re: A border of uniform width is placed around a rectangular [#permalink]

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22 Jan 2016, 01:13

I mean the size of the border ALONE: the question says that the PHOTOGRAPH is 8x10, so it's logical to think that the border has a width bigger than 8 or 10. You said 14x16,but the answer is 3...I don't understand...

I mean the size of the border ALONE: the question says that the PHOTOGRAPH is 8x10, so it's logical to think that the border has a width bigger than 8 or 10. You said 14x16,but the answer is 3...I don't understand...

No, you are confusing width of the border with the final dimensions of photograph + border.

Yes, x=3 is the width but the final dimensions of photo + border are greater than those of photo alone.

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